Class Solution
java.lang.Object
g2001_2100.s2011_final_value_of_variable_after_performing_operations.Solution
2011 - Final Value of Variable After Performing Operations\.
Easy
There is a programming language with only **four** operations and **one** variable `X`:
* `++X` and `X++` **increments** the value of the variable `X` by `1`.
* `--X` and `X--` **decrements** the value of the variable `X` by `1`.
Initially, the value of `X` is `0`.
Given an array of strings `operations` containing a list of operations, return _the **final** value of_ `X` _after performing all the operations_.
**Example 1:**
**Input:** operations = ["--X","X++","X++"]
**Output:** 1
**Explanation:** The operations are performed as follows:
Initially, X = 0.
--X: X is decremented by 1, X = 0 - 1 = -1.
X++: X is incremented by 1, X = -1 + 1 = 0.
X++: X is incremented by 1, X = 0 + 1 = 1.
**Example 2:**
**Input:** operations = ["++X","++X","X++"]
**Output:** 3
**Explanation:** The operations are performed as follows:
Initially, X = 0.
++X: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
X++: X is incremented by 1, X = 2 + 1 = 3.
**Example 3:**
**Input:** operations = ["X++","++X","--X","X--"]
**Output:** 0
**Explanation:** The operations are performed as follows:
Initially, X = 0.
X++: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
--X: X is decremented by 1, X = 2 - 1 = 1.
X--: X is decremented by 1, X = 1 - 1 = 0.
**Constraints:**
* `1 <= operations.length <= 100`
* `operations[i]` will be either `"++X"`, `"X++"`, `"--X"`, or `"X--"`.
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Solution
public Solution()
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finalValueAfterOperations
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